How do you integrate int ( 1-3x)/((2x-1)(x+2)) using partial fractions?

1 Answer
Leland Adriano Alejandro
Jul 19, 2016

color(blue)(int (1-3x)/((2x-1)(x+2))dx= -1/10*ln (2x-1)-7/5*ln(x +2) +C)

Explanation:

From the given int (1-3x)/((2x-1)(x+2))dx
we use partial fraction with the variables A and B
int (1-3x)/((2x-1)(x+2))dx=int A/(2x-1)dx+B/(x +2) dx

Let us determine the constants first
(1-3x)/((2x-1)(x+2))= A/(2x-1)+B/(x +2)
(1-3x)/((2x-1)(x+2))= (A(x+2)+B(2x-1))/((2x-1)(x +2))
(1-3x)/((2x-1)(x+2))= (Ax+2A+2Bx-B)/((2x-1)(x +2))
Now we can set up the equations

A+2B=-3
2A-B=1

Simultaneous solution results to:
A=-1/5
B=-7/5

Now, we can integrate
int (1-3x)/((2x-1)(x+2))dx=int A/(2x-1)dx+B/(x +2) dx
int (1-3x)/((2x-1)(x+2))dx=int (-1/5)/(2x-1)dx+(-7/5)/(x +2) dx
int (1-3x)/((2x-1)(x+2))dx= -1/10*ln (2x-1)-7/5*ln(x +2) +C

God bless ..... I hope the explanation is useful.

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